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This question is related to one that I asked some time ago.


Definition 1. Let $ q \in (-1,1) $. Let $ A $ be a unital $ C^{*} $-algebra and $ (x,y) $ a pair of elements of $ A $. Then define the $ q $-relations for $ (x,y) $ to be the following five relations: \begin{align} x^{*} x + y^{*} y & = 1_{A}. \\ x x^{*} + q^{2} y y^{*} & = 1_{A}. \\ y^{*} y & = y y^{*}. \\ x y & = q y x. \\ x y^{*} & = q y^{*} x. \end{align}

Definition 2. The compact quantum group $ {\text{SU}_{q}}(2) $ is defined as the universal unital $ C^{*} $-algebra that is generated by a pair $ (x,y) $ of elements satisfying the $ q $-relations. We shall call $ (x,y) $ ‘a generating pair for $ {\text{SU}_{q}}(2) $’.

Woronowicz’s proof that all the $ {\text{SU}_{q}}(2) $’s are isomorphic as $ C^{*} $-algebras (contained in his paper Twisted $ \text{SU}(2) $ Group. An Example of a Non-Commutative Differential Calculus) goes as follows:

  1. Let $ q \in (-1,1) $.

  2. Let $ (a,b) $ and $ (\alpha,\beta) $ be generating pairs for $ {\text{SU}_{q}}(2) $ and $ {\text{SU}_{0}}(2) $ respectively.

  3. The pair (of elements of $ {\text{SU}_{q}}(2) $) $$ \phi_{(a,b)} \stackrel{\text{df}}{=} (a ~ f(a^{*} a),b ~ g(b^{*} b)), $$ obtained via the continuous functional calculus, satisfies the $ 0 $-relations, where $ f,g \in C([0,1]) $ and have the following properties: $$ f(t) = \begin{cases} 0 & \text{if $ t = 0 $}, \\ \dfrac{1}{\sqrt{t}} & \text{if $ t \in \left[ 1 - q^{2},1 \right] $}; \end{cases} \qquad g(t) = \begin{cases} 0 & \text{if $ t \in \left[ 0,q^{2} \right] $}, \\ 1 & \text{if $ t = 1 $}. \end{cases} $$ Then by the universal property of $ {\text{SU}_{0}}(2) $, there exists a unique morphism $ \Phi: {\text{SU}_{0}}(2) \to {\text{SU}_{q}}(2) $ such that $ (\Phi(\alpha),\Phi(\beta)) = \phi_{(a,b)} $.

  4. The pair (of elements of $ {\text{SU}_{0}}(2) $) $$ \psi_{(\alpha,\beta)} \stackrel{\text{df}}{=} \left( \sum_{n = 1}^{\infty} \sqrt{1 - q^{2 n}} \left[ (\alpha^{*})^{n - 1} \alpha^{n} - (\alpha^{*})^{n} \alpha^{n + 1} \right], \sum_{n = 0}^{\infty} q^{n} (\alpha^{*})^{n} \beta \alpha^{n} \right) $$ satisfies the $ q $-relations. Then by the universal property of $ {\text{SU}_{q}}(2) $, there exists a unique morphism $ \Psi: {\text{SU}_{q}}(2) \to {\text{SU}_{0}}(2) $ such that $ (\Psi(a),\Psi(b)) = \psi_{(\alpha,\beta)} $.

  5. As $ \Phi \circ \Psi = \text{Id}_{{\text{SU}_{q}}(2)} $ and $ \Psi \circ \Phi = \text{Id}_{{\text{SU}_{0}}(2)} $, one can conclude that $ {\text{SU}_{q}}(2) \cong {\text{SU}_{0}}(2) $.

Step 3 is easily verified via routine continuous functional calculus (the linked post says more about this).

Steps 4 and 5 appear very difficult. I have absolutely no idea how Woronowicz obtained the infinite series in Step 4, but I am very sure that he could not have guessed them. I would therefore like to ask:

Question 1. How can we derive the infinite series in Step 4 and prove that the pair $ \psi_{(\alpha,\beta)} $ satisfies the $ q $-relations? I believe that once this is answered, Step 5 can be tackled.

I must also mention that Woronowicz claims the following identity: $$ \sum_{n = 1}^{\infty} \sqrt{1 - q^{2 n}} \left[ (\alpha^{*})^{n - 1} \alpha^{n} - (\alpha^{*})^{n} \alpha^{n + 1} \right] = \sum_{n = 0}^{\infty} \frac{\left( 1 - q^{2} \right) q^{2 n}}{\sqrt{1 - q^{2 n}} + \sqrt{1 - q^{2 (n + 1)}}} (\alpha^{*})^{n} \alpha^{n + 1}. $$ The infinite series on the right is readily seen to be absolutely convergent. I could easily verify that \begin{align} \forall N \in \Bbb{N}: \qquad & ~ \sum_{n = 1}^{N} \sqrt{1 - q^{2 n}} \left[ (\alpha^{*})^{n - 1} \alpha^{n} - (\alpha^{*})^{n} \alpha^{n + 1} \right] \\ = & ~ \sum_{n = 0}^{N - 1} \frac{\left( 1 - q^{2} \right) q^{2 n}}{\sqrt{1 - q^{2 n}} + \sqrt{1 - q^{2 (n + 1)}}} (\alpha^{*})^{n} \alpha^{n + 1} - \sqrt{1 - q^{2 N}} (\alpha^{*})^{N} \alpha^{N + 1}, \end{align} but I was unable to prove that $ \displaystyle \lim_{N \to \infty} \sqrt{1 - q^{2 N}} (\alpha^{*})^{N} \alpha^{N + 1} = 0_{{\text{SU}_{0}}(2)} $.

Question 2. How can we prove that $ \displaystyle \lim_{N \to \infty} \sqrt{1 - q^{2 N}} (\alpha^{*})^{N} \alpha^{N + 1} = 0_{{\text{SU}_{0}}(2)} $?

These are all the questions that I have. Thank you very much for your help!

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